- #1

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- TL;DR Summary
- As personal curiosity, I want to calculate which is the difference in "travelled height" between a photon that goes across the width of an elevator - which is more or less 2[m] in my country - and a tiny mass particle that free-falls starting at the same "height" as the photon origin, and is released exactly at the same time that the photon is shot from the other side.

As personal curiosity, I want to calculate which is the difference in "travelled height" between a photon that goes across the width of an elevator - which is more or less 2[m] in my country - and a tiny mass particle that free-falls starting at the same "height" as the photon origin, and is released exactly at the same time that the photon is shot from the other side.

I'll make some assumptions in order to make this race "fair enough". Hope I don't make many mistakes and correct me if I'm wrong. We'll see...

First of all, thanks to the sponsors of this "tough" experiment: Coke© light (supplying the laser assembly) and Nanomass© dynamics for the test particle (which they sell by the weight and are dirty cheap).

Also, due the lack of budget, the test will be conducted in Earth, simplifying and considering the

This defines the "Schwarzschild radius", and will be used in the metric:

$$[1]\quad R_s=\frac{2GM}{c^2}$$

The Schwarzschild metric itself:

$$[2]\quad ds^2=-c^2(1-\frac{R_s}{r})dt^2+\frac{dr^2}{(1-\frac{R_s}{r})}+r^2d\theta^2+r^2sin^2(\theta)d\phi^2$$

Both the photon and the test particle will be in "

$$[3]\quad (x^0,x^1,x^2,x^3)=(ct,r,\theta,\phi) \qquad \dot{x}^\mu=\frac{\partial x^\mu}{\partial λ} \qquad \frac{\partial}{\partial λ}(\frac{\partial\mathscr{L}}{\partial\dot{x}^\mu})-\frac{\partial\mathscr{L}}{\partial x^\mu}=0$$

Defining the Lagrangian as the derivative of the line interval by a suitable parameter λ:

$$[4]\quad \mathscr{L}=(\frac{ds}{dλ})^2=-c^2(1-\frac{R_s}{r})\dot{t}^2+\frac{\dot{r}^2}{(1-\frac{R_s}{r})}+r^2\dot{\theta}^2+r^2sin^2(\theta)\dot{\phi}^2$$

Applying the Euler-Lagrange rule to the coordinates, we get the following general conditions that the Geodesic trajectories must follow:

- We can consider the trajectory contained in a plane. Let's choose for sake of simplicity $\theta=\pi/2$:

$$[5]\quad \mathscr{L}(\theta) \Rightarrow 2r^2\ddot{\theta}-2r^2\dot{\theta}^2sin(\theta)cos(\theta)=0 \Rightarrow

\theta=\pi/2$$

- There're a two conserved quantities (constants of motion) to be determined for a Geodesic path:

$$[6]\quad \mathscr{L}(ct) \Rightarrow -2c^2(1-\frac{R_s}{r})\ddot{t}=0 \Rightarrow (1-\frac{R_s}{r})\dot{t}=k$$

$$[7]\quad \mathscr{L}(\phi)\Rightarrow 2r^2sin^2(\theta)\ddot{\phi}=0\Rightarrow r^2sin^2(\theta)\dot{\phi}=l$$

Dividing [7] into [6], we get rid of the derivatives with respect to parameter λ, and can write the constant of motion directly as a consequence of the coordinate components:

$$[9]\quad b=\frac{l}{k}=\frac{r^2\dot{\phi}}{(1-\frac{R_s}{r})\dot{t}}=\frac{r^2}{(1-\frac{R_s}{r})}\frac{d\phi}{dt}$$

Combining all this stuff, we obtain our

$$[10]\quad \Rightarrow \frac{d\phi}{dt}=\frac{1}{r^2}(1-\frac{R_s}{r})\cdot b$$

Now, we have to distinguish between a test particle (with mass), and a (massless) photon.

For light, and using the original metric equation [2]:

$$[11]\quad photon: ds^2=0=-c^2(1-\frac{R_s}{r})dt^2+\frac{dr^2}{(1-\frac{R_s}{r})}+r^2d\phi^2$$

And then, here we have our second equation of motion, related to a constant b to be determined afterwards:

$$[12]\quad (\frac{dr}{dt})^2=-\frac{b^2}{r^2}(1-\frac{R_s}{r})^3+c^2(1-\frac{R_s}{r})^2$$

For the constant of motion, if consider the launch angle ##\varphi## - from the horizontal to the initial direction -, and the primed coordinates (r' and t') for a local observer which always measures (as Einstein likes) the speed of light as c, we have:

$$[13]\quad t=t_0=0,r=r_0 \qquad \frac{dr'}{dt'}|_{t0}=c \cdot sin(\varphi)$$

$$c^2dt'^2=c^2(1-\frac{R_s}{r})dt^2 \qquad dr'^2=\frac{dr^2}{(1-\frac{R_s}{r})}\Rightarrow \frac{dr'}{dt'}=\frac{1}{(1-\frac{R_s}{r})} \cdot \frac{dr}{dt}$$

$$\frac{dr}{dt}|_{t_0}=(1-\frac{R_s}{r_0}) \cdot c \cdot sin(\varphi)$$

$$[14]\quad (1-\frac{R_s}{r_0})^2\cdot c^2 \cdot sin^2(\varphi)=-\frac{b^2}{r_0^2}(1-\frac{R_s}{r_0})^3+c^2(1-\frac{R_s}{r_0})^2$$

$$\Rightarrow b^2=\frac{c^2r_0^2}{(1-\frac{R_s}{r_0})} \cdot \left\{ 1-sin^2(\varphi) \right\}=\frac{c^2r_0^2}{(1-\frac{R_s}{r_0})} \cdot cos^2(\varphi)$$

Now for the test particle. We do the same, except that is not a null but a timelike Geodesic:

$$[15]\quad \mathscr{L}=(\frac{ds}{d\tau})^2=-c^2=-\frac{c^2k^2}{(1-\frac{R_s}{r})}+\frac{(\frac{dr}{d\tau})^2}{(1-\frac{R_s}{r})} \qquad \theta=\pi/2 \quad d\theta=0 \quad d\phi=0$$

$$[16]\quad -c^2(1-\frac{R_s}{r})=-c^2k^2+(\frac{dr}{d\tau})^2$$

$$[17]\quad \frac{dr}{d\tau}=\frac{dr}{dt}\frac{dt}{d\tau}=\frac{dr}{dt}\frac{k}{(1-\frac{R_s}{r})}$$

The second equation of motion for a mass particle:

$$[18]\quad (\frac{dr}{dt})^2=-\frac{c^2}{k^2}(1-\frac{R_s}{r})^3+c^2(1-\frac{R_s}{r})^2$$

From the initial condition, we can get the constant of motion:

$$[19]\quad t=t_0=0,r=r_0 \qquad \frac{dr}{dt}|_{t_0}=0 \qquad \Rightarrow k=±\sqrt{1-\frac{R_s}{r_0}}$$

As for the setup of the test, if we shoot the photon with an initial non-zero launch angle ##\varphi## (inwards), then it will have obviously an initial inward geometric radial velocity, and being c a big number seems unfair to compare then the radial travelled distance of the photon with a poor test particle that starts quiet and still.

So, we'll shot the photon tangentially (##\varphi=0##). This means that if the path of light were unaffected by Gravity, would hit the other side of the elevator at a higher position (i.e., if the starting radius is r_0, would reach the other side at a radius r_1>r_0). In a nutshell, we don't point the laser towards the test particle, but a little above it.

For the test particle, it's starting position will be r_0, and so both the photon and the test particle begin at the same distance (radius) from the center of the planet.

That's fair enough for me.

With a simple Python code (attached), I obtain the following:

FINAL RESULT:

r1-rend=6.556e-16[m] r0-rend_m=2.185e-16[m] tend=6.671e-09

where r1 is the point-of-impact radius of a straight line with a length of 2[m] (we fix as condition the arc covered to be L/r_0), rend if the final radius of the photon position, rend_m is the final radius of the test particle position, and tend is the total flight time (in the coordinate time). So the radial distance displacement for the photon is 3.0004.. times greater than the radial distance displacement of the test particle.

I'll make some assumptions in order to make this race "fair enough". Hope I don't make many mistakes and correct me if I'm wrong. We'll see...

First of all, thanks to the sponsors of this "tough" experiment: Coke© light (supplying the laser assembly) and Nanomass© dynamics for the test particle (which they sell by the weight and are dirty cheap).

Also, due the lack of budget, the test will be conducted in Earth, simplifying and considering the

**Schwarzschild metric**solution to the Einstein Field Equations, so Earth and surrounding spacetime are not flat.This defines the "Schwarzschild radius", and will be used in the metric:

$$[1]\quad R_s=\frac{2GM}{c^2}$$

The Schwarzschild metric itself:

$$[2]\quad ds^2=-c^2(1-\frac{R_s}{r})dt^2+\frac{dr^2}{(1-\frac{R_s}{r})}+r^2d\theta^2+r^2sin^2(\theta)d\phi^2$$

Both the photon and the test particle will be in "

**free-fall**" once released, and thus will follow a**Geodesic**trajectory. To obtain the motion equations, first I'll begin with the**Euler-Lagrange**condition, because such trajectories will maximize some Lagrangian function.$$[3]\quad (x^0,x^1,x^2,x^3)=(ct,r,\theta,\phi) \qquad \dot{x}^\mu=\frac{\partial x^\mu}{\partial λ} \qquad \frac{\partial}{\partial λ}(\frac{\partial\mathscr{L}}{\partial\dot{x}^\mu})-\frac{\partial\mathscr{L}}{\partial x^\mu}=0$$

Defining the Lagrangian as the derivative of the line interval by a suitable parameter λ:

$$[4]\quad \mathscr{L}=(\frac{ds}{dλ})^2=-c^2(1-\frac{R_s}{r})\dot{t}^2+\frac{\dot{r}^2}{(1-\frac{R_s}{r})}+r^2\dot{\theta}^2+r^2sin^2(\theta)\dot{\phi}^2$$

Applying the Euler-Lagrange rule to the coordinates, we get the following general conditions that the Geodesic trajectories must follow:

- We can consider the trajectory contained in a plane. Let's choose for sake of simplicity $\theta=\pi/2$:

$$[5]\quad \mathscr{L}(\theta) \Rightarrow 2r^2\ddot{\theta}-2r^2\dot{\theta}^2sin(\theta)cos(\theta)=0 \Rightarrow

\theta=\pi/2$$

- There're a two conserved quantities (constants of motion) to be determined for a Geodesic path:

$$[6]\quad \mathscr{L}(ct) \Rightarrow -2c^2(1-\frac{R_s}{r})\ddot{t}=0 \Rightarrow (1-\frac{R_s}{r})\dot{t}=k$$

$$[7]\quad \mathscr{L}(\phi)\Rightarrow 2r^2sin^2(\theta)\ddot{\phi}=0\Rightarrow r^2sin^2(\theta)\dot{\phi}=l$$

Dividing [7] into [6], we get rid of the derivatives with respect to parameter λ, and can write the constant of motion directly as a consequence of the coordinate components:

$$[9]\quad b=\frac{l}{k}=\frac{r^2\dot{\phi}}{(1-\frac{R_s}{r})\dot{t}}=\frac{r^2}{(1-\frac{R_s}{r})}\frac{d\phi}{dt}$$

Combining all this stuff, we obtain our

**first general equation of motion**for a Geodesic path:$$[10]\quad \Rightarrow \frac{d\phi}{dt}=\frac{1}{r^2}(1-\frac{R_s}{r})\cdot b$$

Now, we have to distinguish between a test particle (with mass), and a (massless) photon.

For light, and using the original metric equation [2]:

$$[11]\quad photon: ds^2=0=-c^2(1-\frac{R_s}{r})dt^2+\frac{dr^2}{(1-\frac{R_s}{r})}+r^2d\phi^2$$

And then, here we have our second equation of motion, related to a constant b to be determined afterwards:

$$[12]\quad (\frac{dr}{dt})^2=-\frac{b^2}{r^2}(1-\frac{R_s}{r})^3+c^2(1-\frac{R_s}{r})^2$$

For the constant of motion, if consider the launch angle ##\varphi## - from the horizontal to the initial direction -, and the primed coordinates (r' and t') for a local observer which always measures (as Einstein likes) the speed of light as c, we have:

$$[13]\quad t=t_0=0,r=r_0 \qquad \frac{dr'}{dt'}|_{t0}=c \cdot sin(\varphi)$$

$$c^2dt'^2=c^2(1-\frac{R_s}{r})dt^2 \qquad dr'^2=\frac{dr^2}{(1-\frac{R_s}{r})}\Rightarrow \frac{dr'}{dt'}=\frac{1}{(1-\frac{R_s}{r})} \cdot \frac{dr}{dt}$$

$$\frac{dr}{dt}|_{t_0}=(1-\frac{R_s}{r_0}) \cdot c \cdot sin(\varphi)$$

$$[14]\quad (1-\frac{R_s}{r_0})^2\cdot c^2 \cdot sin^2(\varphi)=-\frac{b^2}{r_0^2}(1-\frac{R_s}{r_0})^3+c^2(1-\frac{R_s}{r_0})^2$$

$$\Rightarrow b^2=\frac{c^2r_0^2}{(1-\frac{R_s}{r_0})} \cdot \left\{ 1-sin^2(\varphi) \right\}=\frac{c^2r_0^2}{(1-\frac{R_s}{r_0})} \cdot cos^2(\varphi)$$

Now for the test particle. We do the same, except that is not a null but a timelike Geodesic:

$$[15]\quad \mathscr{L}=(\frac{ds}{d\tau})^2=-c^2=-\frac{c^2k^2}{(1-\frac{R_s}{r})}+\frac{(\frac{dr}{d\tau})^2}{(1-\frac{R_s}{r})} \qquad \theta=\pi/2 \quad d\theta=0 \quad d\phi=0$$

$$[16]\quad -c^2(1-\frac{R_s}{r})=-c^2k^2+(\frac{dr}{d\tau})^2$$

$$[17]\quad \frac{dr}{d\tau}=\frac{dr}{dt}\frac{dt}{d\tau}=\frac{dr}{dt}\frac{k}{(1-\frac{R_s}{r})}$$

The second equation of motion for a mass particle:

$$[18]\quad (\frac{dr}{dt})^2=-\frac{c^2}{k^2}(1-\frac{R_s}{r})^3+c^2(1-\frac{R_s}{r})^2$$

From the initial condition, we can get the constant of motion:

$$[19]\quad t=t_0=0,r=r_0 \qquad \frac{dr}{dt}|_{t_0}=0 \qquad \Rightarrow k=±\sqrt{1-\frac{R_s}{r_0}}$$

As for the setup of the test, if we shoot the photon with an initial non-zero launch angle ##\varphi## (inwards), then it will have obviously an initial inward geometric radial velocity, and being c a big number seems unfair to compare then the radial travelled distance of the photon with a poor test particle that starts quiet and still.

So, we'll shot the photon tangentially (##\varphi=0##). This means that if the path of light were unaffected by Gravity, would hit the other side of the elevator at a higher position (i.e., if the starting radius is r_0, would reach the other side at a radius r_1>r_0). In a nutshell, we don't point the laser towards the test particle, but a little above it.

For the test particle, it's starting position will be r_0, and so both the photon and the test particle begin at the same distance (radius) from the center of the planet.

That's fair enough for me.

With a simple Python code (attached), I obtain the following:

FINAL RESULT:

r1-rend=6.556e-16[m] r0-rend_m=2.185e-16[m] tend=6.671e-09

where r1 is the point-of-impact radius of a straight line with a length of 2[m] (we fix as condition the arc covered to be L/r_0), rend if the final radius of the photon position, rend_m is the final radius of the test particle position, and tend is the total flight time (in the coordinate time). So the radial distance displacement for the photon is 3.0004.. times greater than the radial distance displacement of the test particle.

Python code:

```
from mpmath import *
import numpy as np
import matplotlib.pyplot as plt
import math
from datetime import datetime
mp.dps = 100
mp.pretty = True
c=299792458 # speed of light
c2=pow(c,2) # square of speed of light
G=mpf(6.6743e-11) # Gravitional constant
M=mpf(5.9722e+24) # Earth's mass
r0=mpf(6371000) # Earth's radius
L=mpf(2) # Elevator's width
alp=L/r0
Rs=(2*G*M)/c2
Rs2=pow(Rs,2)
r0s=1-Rs/r0
b2=((pow(r0,2)*c2)/r0s)
b=sqrt(b2)
def zeq(r):
return c2-(b2/pow(r,2))*(1-Rs/r)
def fph_dt_dr(r):
zr=zeq(r)
if (zr>0):
return 1/((1-Rs/r)*sqrt(zr))
else:
raise Exception('NEGATIVE SQRT! r=%5.3e[m]'%(r))
def fph_dphi_dt(r):
return (b/pow(r,2))*(1-Rs/r)
def fph_dphi_dr(r):
zr=zeq(r)
if (zr>0):
return b/sqrt(r*(c2*pow(r,3)-b2*(r-Rs)))
else:
raise Exception('NEGATIVE SQRT! r=%5.3e[m]'%(r))
def fmp_dt_dr(r):
return -1/(c*(1-Rs/r)*sqrt(1-(1-Rs/r)/r0s))
print('')
r1=r0/cos(alp)
[rend] = findroot([lambda r: quad(fph_dphi_dr,[r0,r])-alp],(r1))
tend = quad(fph_dt_dr,[r0,rend])
d=pow(tend,2)*G*M/(2*pow(r0,2)) # initial guess
[rend_m] = findroot([lambda r: quad(fmp_dt_dr,[r0-1e-51,r])-tend],(r0-d))
print('')
print('FINAL RESULT:')
print('r1-rend=%5.3e[m] r0-rend_m=%5.3e[m] tend=%9.3e[s]'%(r1-rend,r0-rend_m,tend))
```